What is equivalent to -36 -8

Health insurance benefits vary by the size of the company (the Henry J. Kaiser Family Foundation website, June 23, 2016). The sa

xxMikexx [17]

Answer:

$\chi^2 = \frac{(32-42)^2}{42}+\frac{(18-8)^2}{8}+\frac{(68-63)^2}{63}+\frac{(7-12)^2}{12}+\frac{(89-84)^2}{84}+\frac{(11-16)^2}{16}=19.221$

Now we can calculate the degrees of freedom for the statistic given by:

$df=(rows-1)(cols-1)=(3-1)(2-1)=2$

And we can calculate the p value given by:

$p_v = P(\chi^2_{2} >19.221)=0.000067$

And we can find the p value using the following excel code:

"=1-CHISQ.DIST(19.221,2,TRUE)"

Since the p values is higher than a significance level for example $\alpha=0.05$ , we can reject the null hypothesis at 5% of significance, and we can conclude that the two variables are dependent at 5% of significance.

Step-by-step explanation:

Previous concepts

A chi-square goodness of fit test "determines if a sample data matches a population".

A chi-square test for independence "compares two variables in a contingency table to see if they are related. In a more general sense, it tests to see whether distributions of categorical variables differ from each another".

Solution to the problem

Assume the following dataset:

Size Company/ Heal. Ins. Yes No Total

Small 32 18 50

Medium 68 7 75

Large 89 11 100

_____________________________________

Total 189 36 225

We need to conduct a chi square test in order to check the following hypothesis:

H0: independence between heath insurance coverage and size of the company

H1: NO independence between heath insurance coverage and size of the company

The statistic to check the hypothesis is given by:

$\sum_{i=1}^n \frac{(O_i -E_i)^2}{E_i}$

The table given represent the observed values, we just need to calculate the expected values with the following formula $E_i = \frac{total col * total row}{grand total}$

And the calculations are given by:

$E_{1} =\frac{50*189}{225}=42$

$E_{2} =\frac{50*36}{225}=8$

$E_{3} =\frac{75*189}{225}=63$

$E_{4} =\frac{75*36}{225}=12$

$E_{5} =\frac{100*189}{225}=84$

$E_{6} =\frac{100*36}{225}=16$

And the expected values are given by:

Size Company/ Heal. Ins. Yes No Total

Small 42 8 50

Medium 63 12 75

Large 84 16 100

_____________________________________

Total 189 36 225

And now we can calculate the statistic:

$\chi^2 = \frac{(32-42)^2}{42}+\frac{(18-8)^2}{8}+\frac{(68-63)^2}{63}+\frac{(7-12)^2}{12}+\frac{(89-84)^2}{84}+\frac{(11-16)^2}{16}=19.221$

Now we can calculate the degrees of freedom for the statistic given by:

$df=(rows-1)(cols-1)=(3-1)(2-1)=2$

And we can calculate the p value given by:

$p_v = P(\chi^2_{2} >19.221)=0.000067$

And we can find the p value using the following excel code:

"=1-CHISQ.DIST(19.221,2,TRUE)"

Since the p values is higher than a significance level for example $\alpha=0.05$ , we can reject the null hypothesis at 5% of significance, and we can conclude that the two variables are dependent at 5% of significance.

3 0

3 years ago

A line passes through the points (2,21) and (8,27). Write a linear function rule in terms of x and y for

tatiyna

We have two points so we can find the gradient using y1-y2/x1-x2
gradient = 21-27/2-8
= 1
we know the form for any linear equation is y = mx + c
we have m and a point so we can substitute in point (2,21) to find c
21 = 1 x 2 + c
c = 19
therefore, the equation is y = x + 19

6 0

2 years ago

Read 2 more answers

Two different simple random samples are drawn from two different populations. The first sample consists of 30 people with 16 hav

Furkat [3]

Answer:

There is no significant evidence that p1 is different than p2 at 0.01 significance level.

99% confidence interval for p1-p2 is -0.171 ±0.237 that is (−0.408, 0.066)

Step-by-step explanation:

Let p1 be the proportion of the common attribute in population1

And p2 be the proportion of the same common attribute in population2

$H_{0}$ : p1-p2=0

$H_{a}$ : p1-p2≠0

Test statistic can be found using the equation:

$z=\frac{p2-p1}{\sqrt{{p*(1-p)*(\frac{1}{n1} +\frac{1}{n2}) }}}$ where

p1 is the sample proportion of the common attribute in population1 ( $\frac{16}{30} =0.533$ )

p2 is the sample proportion of the common attribute in population2 ( $\frac{1337}{1900} =0.704$ )

p is the pool proportion of p1 and p2 ( $\frac{16+1337}{30+1900}=0.701$ )

n1 is the sample size of the people from population1 (30)

n2 is the sample size of the people from population2 (1900)

Then $z=\frac{0.704-0.533}{\sqrt{{0.701*0.299*(\frac{1}{30} +\frac{1}{1900}) }}}$ ≈ 2.03

p-value of the test statistic is 0.042>0.01, therefore we fail to reject the null hypothesis. There is no significant evidence that p1 is different than p2.

99% confidence interval estimate for p1-p2 can be calculated using the equation

p1-p2± $z*\sqrt{\frac{p1*(1-p1)}{n1}+\frac{p2*(1-p2)}{n2}}$ where